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Among the following molecules/ions, C(2...

Among the following molecules/ions, `C_(2)^(2-), N_(2)^(2-), O_(2)^(2-), O_(2)`
Which one is diamagnetic and has the shortest bond length?

A

`C_(2)^(2-)`

B

`O_(2)^(2-)`

C

`O_(2)`

D

`N_(2)^(2-)`

Text Solution

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The correct Answer is:
To solve the question, we need to analyze the given molecules/ions: \(C_2^{2-}\), \(N_2^{2-}\), \(O_2^{2-}\), and \(O_2\). We will determine which of these is diamagnetic and has the shortest bond length. ### Step-by-Step Solution: 1. **Determine the Total Number of Electrons for Each Species**: - \(C_2^{2-}\): Each carbon has 6 electrons, so \(2 \times 6 + 2 = 14\) electrons. - \(N_2^{2-}\): Each nitrogen has 7 electrons, so \(2 \times 7 + 2 = 16\) electrons. - \(O_2^{2-}\): Each oxygen has 8 electrons, so \(2 \times 8 + 2 = 18\) electrons. - \(O_2\): Each oxygen has 8 electrons, so \(2 \times 8 = 16\) electrons. 2. **Fill the Molecular Orbitals Using Molecular Orbital Theory (MOT)**: - **For \(C_2^{2-}\)**: - Molecular Orbital Filling: \(\sigma_{1s}^2 \sigma_{1s}^* \sigma_{2s}^2 \sigma_{2s}^* \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2\) - Total Bonding Electrons = 10, Anti-bonding Electrons = 4 - Bond Order = \(\frac{10 - 4}{2} = 3\) - Unpaired Electrons = 0 (Diamagnetic) - **For \(N_2^{2-}\)**: - Molecular Orbital Filling: \(\sigma_{1s}^2 \sigma_{1s}^* \sigma_{2s}^2 \sigma_{2s}^* \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^* \pi_{2p_y}^*\) - Total Bonding Electrons = 10, Anti-bonding Electrons = 6 - Bond Order = \(\frac{10 - 6}{2} = 2\) - Unpaired Electrons = 2 (Paramagnetic) - **For \(O_2^{2-}\)**: - Molecular Orbital Filling: \(\sigma_{1s}^2 \sigma_{1s}^* \sigma_{2s}^2 \sigma_{2s}^* \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^* \pi_{2p_y}^*\) - Total Bonding Electrons = 10, Anti-bonding Electrons = 8 - Bond Order = \(\frac{10 - 8}{2} = 1\) - Unpaired Electrons = 2 (Paramagnetic) - **For \(O_2\)**: - Molecular Orbital Filling: \(\sigma_{1s}^2 \sigma_{1s}^* \sigma_{2s}^2 \sigma_{2s}^* \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2\) - Total Bonding Electrons = 10, Anti-bonding Electrons = 6 - Bond Order = \(\frac{10 - 6}{2} = 2\) - Unpaired Electrons = 2 (Paramagnetic) 3. **Identify the Diamagnetic Species**: - The only diamagnetic species among the given options is \(C_2^{2-}\) since it has no unpaired electrons. 4. **Determine the Shortest Bond Length**: - Bond length is inversely related to bond order. The higher the bond order, the shorter the bond length. - Bond Order Comparison: - \(C_2^{2-}\): Bond Order = 3 - \(N_2^{2-}\): Bond Order = 2 - \(O_2^{2-}\): Bond Order = 1 - \(O_2\): Bond Order = 2 - Since \(C_2^{2-}\) has the highest bond order (3), it will have the shortest bond length. ### Conclusion: The molecule that is diamagnetic and has the shortest bond length is **\(C_2^{2-}\)**. ---
To solve the question, we need to analyze the given molecules/ions: \(C_2^{2-}\), \(N_2^{2-}\), \(O_2^{2-}\), and \(O_2\). We will determine which of these is diamagnetic and has the shortest bond length. ### Step-by-Step Solution: 1. **Determine the Total Number of Electrons for Each Species**: - \(C_2^{2-}\): Each carbon has 6 electrons, so \(2 \times 6 + 2 = 14\) electrons. - \(N_2^{2-}\): Each nitrogen has 7 electrons, so \(2 \times 7 + 2 = 16\) electrons. - \(O_2^{2-}\): Each oxygen has 8 electrons, so \(2 \times 8 + 2 = 18\) electrons. ...
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